Question

In the figure above, points A, D, B, and G are collinear. If $\angle CAD$ measures ${76}^{\circ }$, $\angle BCD$ measures ${47}^{\circ }$, and $\angle CBG$ measures ${140}^{\circ }$, find the degree measure of $\angle ACD$.

• A. ${12}^{\circ }$
• B. ${14}^{\circ }$
• C. ${17}^{\circ }$
• D. ${36}^{\circ }$
• E. ${43}^{\circ }$

Answer 1

Answer: (C)

${17}^{\circ }$

As points $A,D,B$ and $G$ are collinear

$\therefore \angle CBD+\angle CBG={180}^{\circ }$

$\Rightarrow \angle CBD+{140}^{\circ }={180}^{\circ }$

$\Rightarrow \angle CBD={180}^{\circ }-{140}^{\circ }$

In $△CBD$

$\Rightarrow \angle DCB+\angle CDB+\angle CBD={180}^{\circ }$

$\Rightarrow {47}^{\circ }+\angle CBD+{40}^{\circ }={180}^{\circ }$

$\Rightarrow \angle CDB={180}^{\circ }-({47}^{\circ }+{40}^{\circ })$

$\Rightarrow \angle CDB={180}^{\circ }-{87}^{\circ }\Rightarrow {93}^{\circ }$

Point $D$ is lies in line $AB$

$\therefore \angle ADC+\angle CDB={180}^{\circ }$

$\Rightarrow \angle ADC+{93}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ADC={180}^{\circ }-{93}^{\circ }$

$\Rightarrow \angle ADC={87}^{\circ }$

In $△CAD$

$\Rightarrow \angle ACD+\angle CDA+\angle CDA={180}^{\circ }$

$\Rightarrow \angle ACD+{87}^{\circ }+{76}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ACD+{163}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ACD={180}^{\circ }-{163}^{\circ }\Rightarrow {17}^{\circ }$